The equation 2^|x^2-12|=√(e^|x|log 4) has

We have,

$2^{|x^2-12|}=\sqrt{e^{|x|\log 4}}$

$⇒2^{|x^2-12|}=\sqrt{e^{\log 4^{|x|}}}$

$⇒2^{|x^2-12|}=\sqrt{4^{|x|}}$

$⇒2^{|x^2-12|}=2^{|x|}⇒|x^2-12|=|x|$

Now, $x^2-12=0⇒ x=±2\sqrt{3}$. So, we have the following cases.

CASE I When $x ≥2\sqrt{3}$

In this case, we have

$|x^2-12|=x^2-12$ and $|x|= x$

$∴|x^2-12|=|x|$

$⇒x^2-12=x$

$⇒x^2-x-12=0$

$⇒(x-4) (x+3)=0⇒ x=4$  $[∵x≥2\sqrt{3}]$

CASE II When $x ≤-2\sqrt{3}$

In this case, we have

$|x^2-12|=x^2-12$ and $|x|= x$

$∴|x^2-12|=|x|$

$⇒x^2-12=-x$

$⇒x^2+x-12=0$

$⇒(x+4) (x-3)=0⇒ x=-4$  $[∵x≥-2\sqrt{3}]$

CASE III When $-2\sqrt{3}<x<0$

In this case, we have

$|x^2-12|=-(x^2-12)$ and $|x|= x$

$∴|x^2-12|=|x|$

$⇒-(x^2-12)=-x$

$⇒x^2-x-12=0$

$⇒(x-4) (x+3)=0⇒ x=-3$  $[∵-2\sqrt{3}<x<0]$

CASE IV When $0≤x<2\sqrt{3}$

In this case, we have

$|x^2-12|=-(x^2-12)$ and $|x|= x$

$∴|x^2-12|=|x|$

$⇒-(x^2-12)=x$

$⇒x^2+x-12=0$

$⇒(x+4) (x-3)=0⇒ x=3$  $[∵0≤x<2\sqrt{3}]$

Hence, the four real solutions are ± 4, ± 3. Clearly, their sum is zero.